Counter for items in an inventory using iterators - c++

Question

I'm somewhat new to vectors and iterators and I'm trying to figure out how to display an amount of items using an iterator. One example would be you have 5 apples. I would want it to output "5x Apple" or something of that sort. I have no idea how one might accomplish this. Here's a simple code that has the user put in a string to add to the inventory.
#include <iostream>
#include <string>
#include <vector>
using namespace std;
string item;
vector<string> inventory;
vector<string>::iterator iter;
int main()
{
while(true){
cin >> item;
inventory.push_back(item);
cout << "INVENTORY:\n";
for(iter = inventory.begin(); iter != inventory.end(); iter++)
cout << *iter << endl;
}
}
EDIT: I'm trying to make an inventory system for a game. That's why I thought i may need an iterator. If there's a better way to make an inventory system than using an iterator, please let me know. Apologies, i should have clarified.

Answer 1

An iterator lets you iterate through a container, but it does not do any counting for you.
A container's size() tells you how many items are in the container, but if you have different types of items then you have to count them yourself.
For instance, say you have 4 "apple" and 1 "orange".
You have to look at each item entered and count it as needed, eg:
#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<string> inventory;
int numApples = 0;
int numOranges = 0;
int numOther = 0;
int main()
{
string item;
while (cin >> item)
{
inventory.push_back(item);
if (item == "apples")
++numApples;
else if (item == "orange")
++numOranges;
else
++numOther;
}
cout << "INVENTORY:\n";
for (vector<string>::iterator iter = inventory.begin(); iter != inventory.end(); ++iter)
cout << *iter << endl;
/* or, if you are using C++11 or later:
for (string &s : inventory)
cout << s << endl;
*/
cout << "# apples: " << numApples << endl;
cout << "# oranges: " << numOranges << endl;
cout << "# other: " << numOther << endl;
return 0;
}
Or, you might consider using std::count_if(), eg:
#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<string> inventory;
bool isApple(const string &s) { return (s == "apple"); }
bool isOrange(const string &s) { return (s == "orange"); }
bool isOther(const string &s) { return !(isApple(s) || isOrange(s)); }
int main()
{
string item;
while (cin >> item)
inventory.push_back(item);
cout << "INVENTORY:\n";
for (vector<string>::iterator iter = inventory.begin(); iter != inventory.end(); ++iter)
cout << *iter << endl;
/* or, if you are using C++11 or later:
for (string &s : inventory)
cout << s << endl;
*/
cout << "# apples: " << count_if(inventory.begin(), inventory.end(), isApple) << endl;
cout << "# oranges: " << count_if(inventory.begin(), inventory.end(), isOrange) << endl;
cout << "# other: " << count_if(inventory.begin(), inventory.end(), isOther) << endl;
/* or, if you are using C++11 or later:
cout << "# apples: " << count_if(inventory.begin(), inventory.end(), [](auto &s){ return (s == "apple"); }) << endl;
cout << "# oranges: " << count_if(inventory.begin(), inventory.end(), [](auto &s){ return (s == "orange"); }) << endl;
cout << "# other: " << count_if(inventory.begin(), inventory.end(), [](auto &s){ return (s != "apple") && (s != "orange"); }) << endl;
*/
return 0;
}
Update: based on another question you posted, try something more like this instead:
#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<string> other_inventory;
int numApples = 0;
int numOranges = 0;
int main()
{
string item;
while (cin >> item)
{
if (item == "apples")
++numApples;
else if (item == "orange")
++numOranges;
else
other_inventory.push_back(item);
}
cout << "INVENTORY:\n";
if (numApples > 0)
cout << "# apples: " << numApples << endl;
if (numOranges > 0)
cout << "# oranges: " << numOranges << endl;
for (vector<string>::iterator iter = other_inventory.begin(); iter != other_inventory.end(); ++iter)
cout << *iter << endl;
/* or, if you are using C++11 or later:
for (string &s : other_inventory)
cout << s << endl;
*/
return 0;
}

Answer 2

A way to simplify this is to sort the items in the inventory. That will bring the identical ones together, which simplifies counting them. Start at the beginning, count the number of consecutive items that match the current one, display it, and continue with the first non-matching one.
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
int main()
{
std::string item;
std::vector<std::string> inventory;
while (true) {
std::cin >> item;
if (item == "quit") return 0;
inventory.push_back(item);
std::sort(inventory.begin(), inventory.end());
std::cout << "INVENTORY:\n";
auto current = inventory.begin();
while (current != inventory.end()) {
int count = 1;
auto probe = current + 1;
while (probe != inventory.end() && *probe == *current) {
++count;
++probe;
}
std::cout << count << "x " << *current << '\n';
current = probe;
}
std::cout.flush();
}
return 0;
}
In more detail, if your inventory is {"orange", "apple", "orange"}, then the sort will rearrange the order to {"apple", "orange", "orange"}. Note that the identical ones are together.
Now the iterator current starts at the beginning ("apple"). We set count to 1 because we know there's at least 1. We set the iterator probe to point to the next item ("orange"). Since the value at probe doesn't match the value at current, the inner loop does nothing. We print count and the current item ("apple"). We continue by setting current to probe, because, at this point, probe will point to the first item that didn't match the current one.
On the second iteration, current refers to the first "orange". We reset count to 1 and start probe at the next item ("orange"). Since the values at match, we increment count (now 2) and advance probe (now at the end of the inventory). We print count (2) and the current item ("orange"), and set current to probe (end of the list). The outer loop condition sees we're at the end of the inventory, so the loop terminates.

Answer 3

inventory.size()
returns the number of items in your vector.
I don't see how you would need an iterator for that task.